Optimization with inequality constraints

Consider the problem max 𝑥 1 ,𝑥 2 𝑢 𝑥 1 , 𝑥 𝑠. 𝑡. 𝑦 + 𝑥 ≤ 4 2 𝑦 + 2𝑥 ≤ 6 2

Optimization with inequality constraints: the Kuhn-Tucker (KT) conditions

The

KT conditions

for the problem max

x f

(

x

) subject to

g j

(

x

) ≤

c j

for

j

= 1, ...,

m

are L

i

'(

x

) = 0 for

i

= 1 ,...,

n

λ

j

≥ 0, where

g j

(

x

) ≤

c j

and λ

j

[

g j

(

x

) −

c j

] = 0 for

j

= 1, ...,

m

.

𝐿(𝑥) = 𝑓 (𝑥) − 𝑚 𝑗=1 𝜆 𝑗 (𝑔 𝑗 (𝑥) − 𝑐 𝑗 ) .

Example max {x 1 ,x 2 } −(x 1 − 4) 2 − (x x 1 x 1 s. t.

+ x 2 + 3x 2 ≤ 4 ≤ 9 2 − 4) 2 𝐿 𝑥 = − x 1 − 4 2 − x 2 − 4 2 − 𝜆 1 (x 1 + x 2 − 4) − 𝜆 2 (x 1 + 3x 2 − 9) Kuhn Tucker conditions are −2(

x

1 − 4) − λ 1 − λ 2 = 0 −2(

x

2 − 4) − λ 1 − 3λ 2 = 0

x

1 +

x

2 ≤ 4, λ 1 ≥ 0, and

x

1 + 3

x

2 ≤ 9, λ 2 ≥ 0, and λ 1 (

x

1 +

x

2 − 4)= 0 λ 2 (

x

1 + 3

x

2 − 9)= 0 4

When KT conditions are necessary

Let 𝑓 and 𝑔 𝑗 for 𝑗 = 1, … , 𝑚 be continuously differentiable functions of many variables and let constants. Suppose that 𝑥 ∗ 𝑐 𝑗 for solves the problem 𝑗 = 1, … , 𝑚 be max 𝑓 𝑥 𝑠. 𝑡. 𝑔 𝑗 (𝑥) ≤ 𝑐 𝑗 𝑓𝑜𝑟 𝑗 = 1, … , 𝑚 .

Suppose that - either each 𝑔 𝑗 is concave - or each 𝑔 𝑗 is convex and there is some 𝑔 𝑗 (𝑥) < 𝑐 𝑗 for 𝑗 = 1, … , 𝑚 𝑥 such that - or each 𝑔 𝑗 is quasiconvex, 𝛻 𝑔 𝑗 (𝑥 ∗ ) ≠ (0, … , 0) ∀𝑗 , and there is some 𝑥 such that 𝑔 𝑗 (𝑥) < 𝑐 𝑗 for 𝑗 = 1, … , 𝑚 .

Then there exists a unique vector 𝜆 = (𝜆 1 , … , 𝜆 𝑚 ) (𝑥 ∗ , 𝜆) satisfies the Kuhn-Tucker conditions such that 5

Example: KT are not necessary conditions for a max max 𝑥,𝑦 𝑥 𝑠. 𝑡.

y−(1−x)

3

≤ 0 and y ≥ 0

The constraint does not satisfy any of the conditions in the proposition. Indeed consider the first constraint 𝐽 = 3(1 − 𝑥) 2 1 𝐻 = −6(1 − 𝑥) 0 0 0 𝐻 𝑏 = 0 3(1 − 𝑥) 2 1 3(1 − 𝑥) −6(1 − 𝑥) 0 2 1 0 0 Then the constraint is not concave, convex or quasiconvex Quasiconcavity: Slides 36-37 lezione precedente 6

The solution is 𝑥 = 1 𝑦 = 0 0,9 0,7 0,5 0,3 y C1 0,1 -0,1 0 0,2 0,4 0,6 0,8 1 1,2 1,4 1,6 1,8 2 -0,3 -0,5 7

The Lagrangean is 𝐿(𝑥) = 𝑥 − 𝜆 1 (𝑦 − (1 − 𝑥) 3 ) + 𝜆 2 𝑦.

The Kuhn-Tucker conditions are 1 − 3𝜆 1 (1 − 𝑥) 2 = 0 −𝜆 1 + 𝜆 2 = 0 𝑦 − (1 − 𝑥) 3 ≤ 0 , 𝜆 1 ≥ 0 , and 𝜆 1 𝑦 − (1 − 𝑥) 3 = 0 −𝑦 ≤ 0 , 𝜆 2 ≥ 0 , and 𝜆 2 −𝑦 = 0 .

These conditions have no solution. From the last condition, either 𝜆 2 = 0 or 𝑦 = 0 . If 𝜆 2 = 0 then 𝜆 1 = 0 from the second condition, so that no value of 𝑥 is compatible with the first condition. If 𝑦 = 0 then from the third condition either 𝜆 1 = 0 or 𝑥 = 1 , both of which are incompatible with the first condition.

the sufficiency of the Kuhn-Tucker conditions (1)

Let 𝑓 and 𝑔 𝑗 for 𝑗 = 1, … , 𝑚 be continuously differentiable functions of many variables and let 𝑐 𝑗 for 𝑗 = 1, … , 𝑚 be constants. Consider the problem max 𝑥 𝑓 𝑥 𝑠. 𝑡. 𝑔 Suppose that 𝑗 ≤ 𝑐 𝑗 for 𝑗 = 1, … , 𝑚 .

- 𝑓 is concave and - 𝑔 𝑗 is quasiconvex for 𝑗 = 1, … , 𝑚 .

If there exists 𝜆 = (𝜆 1 , … , 𝜆 𝑚 ) such that (𝑥 ∗ , 𝜆) satisfies the Kuhn-Tucker conditions then 𝑥 ∗ solves the problem 9

the sufficiency of the Kuhn-Tucker conditions (2)

Let 𝑓 and 𝑔 𝑗 for 𝑗 = 1, … , 𝑚 be continuously differentiable functions of many variables and let 𝑐 𝑗 for 𝑗 = 1, … , 𝑚 be constants. Consider the problem max 𝑥 𝑓 𝑥 𝑠. 𝑡. 𝑔 Suppose that 𝑗 ≤ 𝑐 𝑗 for 𝑗 = 1, … , 𝑚 .

- 𝑓 is twice differentiable and quasiconcave and - 𝑔 𝑗 is quasiconvex for 𝑗 = 1, … , 𝑚 .

If there exists 𝜆 = (𝜆 1 , … , 𝜆 𝑚 ) and a value of 𝑥 ∗ such that (𝑥 ∗ , 𝜆) satisfies the Kuhn-Tucker conditions and 𝑓′ 𝑖 (𝑥 ∗ ) ≠ 0 for 𝑖 = 1, … , 𝑛 then 𝑥 ∗ solves the problem.

10

Necessity and sufficiency of KT conditions

A) The KT conditions are both necessary and sufficient –

if

the objective function is concave and –

either

each constraint is linear –

or

each constraint function is convex and some vector of the variables satisfies all constraints strictly.

Necessity and sufficiency of KT conditions

B) Suppose that - the objective quasiconcave and function is twice differentiable and - every constraint is linear.

Then - If

x

* solves the problem then there exists a unique vector λ such that (

x

*, λ) satisfies the Kuhn-Tucker conditions, and - if (

x

*, λ) satisfies the Kuhn-Tucker conditions and

f i

' (

x

*) ≠ 0 for

i

= 1, ...,

n

then

x

* solves the problem.

max {𝑥 1 ,𝑥 2 } [−(𝑥 1

Example

− 4) 2 − (𝑥 2 − 4) 2 ] 𝑥 1 𝑥 1 𝑠. 𝑡.

+ 𝑥 2 + 3𝑥 2 ≤ 4 ≤ 9 The objective function is concave and the constraints are both linear, so the solutions of the problem are the solutions of the Kuhn-Tucker conditions.

13

Kuhn Tucker conditions are −2(

x

1 − 4) − λ 1 − λ 2 = 0 −2(

x

2 − 4) − λ 1 − 3λ 2 = 0

x

1 +

x

2 ≤ 4, λ 1 ≥ 0, and

x

1 + 3

x

2 ≤ 9, λ 2 ≥ 0, and λ 1 (

x

1 +

x

2 − 4)= 0 λ 2 (

x

1 + 3

x

2 − 9)= 0 To solve this system of condition we have to consider all possibilities about the values of lambdas We have to consider the following 4 cases: 1) λ 1 = λ 2 = 0 2) λ 1 >0 λ 2 = 0 3) λ 1 =0 λ 2 > 0 4) λ 1 >0 λ 2 > 0 14

Kuhn Tucker conditions are −2(

x

1 −2(

x

2

x x

1 1 − 4) − λ 1 − 4) − λ 1 − λ 2 − 3λ 2 = 0 = 0 +

x

2 + 3

x

2 ≤ 4, λ 1 ≤ 9, λ 2 ≥ 0, and λ 1 (

x

1 ≥ 0, and λ 2 (

x

1 +

x

2 − 4)= 0 + 3

x

2 − 9)= 0

Case 1: λ

1

= λ

2

= 0

KT conditions are −2(

x

1 −2(

x

2

x x

1 1 + − 4) = 0 − 4) = 0

x

2 ≤ 4, + 3

x

2 ≤ 9 Then

x

1 = 4 and

x

2 =4 It not a solution because the last two inequalities are not satisfied 15

Kuhn Tucker conditions are −2(

x

1 −2(

x

2

x x

1 1 − 4) − λ 1 − 4) − λ 1 − λ 2 − 3λ 2 = 0 = 0 +

x

2 + 3

x

2 ≤ 4, λ 1 ≤ 9, λ 2 ≥ 0, and λ 1 (

x

1 ≥ 0, and λ 2 (

x

1 +

x

2 − 4)= 0 + 3

x

2 − 9)= 0

Case 2: λ

1

>0 λ

2

= 0

KT conditions are −2(

x

1 −2(

x

2

x

1 +

x

2 − 4) − λ 1 − 4) − λ 1 − 4= 0

x

1 + 3

x

2 ≤ 9, = 0 = 0 From the first 2 equations

x

1 =

x

2 Using the third equation we get

x

1 =

x

2 =2 and λ 1 =4 It is a solution because the last inequality is satisfied 16

Kuhn Tucker conditions are −2(

x

1 −2(

x

2

x x

1 1 − 4) − λ 1 − 4) − λ 1 − λ 2 − 3λ 2 = 0 = 0 +

x

2 + 3

x

2 ≤ 4, λ 1 ≤ 9, λ 2 ≥ 0, and λ 1 (

x

1 ≥ 0, and λ 2 (

x

1 +

x

2 − 4)= 0 + 3

x

2 − 9)= 0

Case 3: λ

1

=0 λ

2

> 0

KT conditions are −2(

x

1 −2(

x

2

x

1 +

x

2 − 4) − λ 2 − 4) − 3λ 2 = 0 = 0 ≤ 4

x

1 + 3

x

2 − 9= 0 From the first 2 equations

x

2 =3

x

1 -8 Using the last equation we get

x

1 = 3.3

It is not a solution because it does not satisfy the inequality 17

Kuhn Tucker conditions are −2(

x

1 −2(

x

2

x x

1 1 − 4) − λ 1 − 4) − λ 1 − λ 2 − 3λ 2 = 0 = 0 +

x

2 + 3

x

2 ≤ 4, λ 1 ≤ 9, λ 2 ≥ 0, and λ 1 (

x

1 ≥ 0, and λ 2 (

x

1 +

x

2 − 4)= 0 + 3

x

2 − 9)= 0

Case 4: λ

1

>0 λ

2

> 0

KT conditions are −2(

x

1 − 4) − λ 1 − λ 2 = 0 −2(

x

2 − 4) − λ 1 − 3λ 2 = 0

x

1 +

x

2 − 4= 0

x

1 + 3

x

2 = 0 18

KT conditions are −2(

x

1 − 4) − λ 1 − λ 2 = 0 −2(

x

2 − 4) − λ 1 − 3λ 2 = 0

x

1 +

x

2 − 4= 0

x

1 + 3

x

2 = 0 Using the last two equation we get

x

1 =1.5 and

x

2 =2.5

Replacing in the first two equation we get the values of lambdas λ 1 =6 λ 2 = - 1 This is not a solution because it violates the condition λ 2 ≥ 0. 19

Solution is

x

1 =

x

2 =2 and λ 1 =4 20

Optimization with inequality constraints: non negativity constraints

The general form of such a problem is: max

x f

(

x

) subject to

g j

(

x

) ≤

c j

for

j

= 1, ...,

m

and

x i

≥ 0 for

i

= 1, ...,

n

.

Lagrangean is 𝐿 𝑥 = 𝑓 𝑥 − 𝑚 𝑗=1 𝜆 𝑗 𝑔 𝑗 𝑥 − 𝑐 𝑗 − 𝑛 𝑗=1 𝜆 𝑚 + 𝑗 (−𝑥 𝑗 ) It is a special case of the general maximization problem with inequality constraints: the nonnegativity constraint on each variable is simply an additional inequality constraint.

Specifically, if we define the function

g m

+

i

for

i

= 1, ...,

n

by

g m

+

i

(

x

) = −

x i

and let

c m

+

i

= 0 for

i

= 1, ...,

n

, then we may write the problem as max

x f

(

x

) subject to

g j

(

x

) ≤

c j

for

j

= 1, ...,

m

+

n

and solve it using the Kuhn-Tucker conditions

Optimization with inequality constraints: non negativity constraints

Approaching the problem in this way involves working with

n

+

m

Lagrange multipliers, which can be difficult if

n

is large.

Then we can use an alternative approach, the

Lagrangean modified

Consider the following problem: max

x f

(

x

) subject to

g j

(

x

) ≤

c j

for

j

= 1, ...,

m

and

x i

≥ 0 for

i

= 1, ...,

n

.

The

modified Lagrangean is:

𝑀(𝑥) = 𝑓 (𝑥) − 𝑚 𝑗=1 𝜆 𝑗 (𝑔 𝑗 (𝑥) − 𝑐 𝑗 )

The

modified Lagrangean is:

𝑀(𝑥) = 𝑓 (𝑥) − 𝑚 𝑗=1 𝜆 𝑗 (𝑔 𝑗 (𝑥) − 𝑐 𝑗 ) Kuhn-Tucker conditions for the modified Lagrangean: 𝑀 𝑖 ′(𝑥) ≤ 0 𝑔 𝑗 (𝑥) ≤ 𝑐 𝑗 , , 𝜆 𝑗 𝑥 𝑖 ≥ 0 ≥ 0 and and 𝑥 𝑖 · 𝑀𝑖′(𝑥) = 0 for 𝑖 = 1, … , 𝑛 𝜆 𝑗 · [𝑔 𝑗 𝑥 − 𝑐 𝑗 ] = 0 for 𝑗 = 1, . . , 𝑚

in any problem for which the original Kuhn-Tucker conditions may be used, we may alternatively use the conditions for the modified Lagrangean. For most problems in which the variables are constrained to be nonnegative, the Kuhn-Tucker conditions for the modified Lagrangean are easier than the conditions for the original Lagrangean Example.

Consider the problem max

x

,

y xy

subject to

x

+

y

≤ 6,

x

≥ 0, and

y

≥ 0

Function xy is twice-differentiable and quasiconcave and the constraint functions are linear, so the Kuhn-Tucker conditions are necessary and if ((

x

*,

y

*), λ*) satisfies these conditions and no partial derivative of the objective function at (

x

*,

y

*) is zero then (

x

*,

y

*) solves the problem.

Solutions of the Kuhn-Tucker conditions at which all derivatives of the objective function are zero may or may not be solutions of the problem We try to solve it 1) using the lagrangean 2) Using the modified lagrangean

1) Using Lagrangean

𝐿 𝑥, 𝑦 = 𝑥𝑦 − 𝜆 1 (𝑥 + 𝑦 − 6) − 𝜆 2 (−𝑥) − 𝜆 3 (−𝑦) Kuhn Tucker conditions are: 𝑦 − 𝜆 1 + 𝜆 2 = 0 𝑥 − 𝜆 1 + 𝜆 3 = 0 𝜆 1 ≥ 0, 𝑥 + 𝑦 ≤ 6, 𝜆 1 𝑥 + 𝑦 − 6 = 0 𝜆 2 ≥ 0, −𝑥 ≤ 0, 𝜆 2 −𝑥 = 0 𝜆 3 ≥ 0, −𝑦 ≤ 0, 𝜆 3 −𝑦 = 0 27

We have to consider the following 8 cases: 1) λ 1 =0 λ 2 = 0 λ 3 = 0 2) λ 1 >0 λ 2 = 0 λ 3 = 0 3) λ 1 =0 λ 2 > 0 λ 3 = 0 4) λ 1 >0 λ 2 > 0 λ 3 = 0 5) λ 1 =0 λ 2 = 0 λ 3 > 0 6) λ 1 >0 λ 2 = 0 λ 3 > 0 7) λ 1 =0 λ 2 > 0 λ 3 > 0 8) λ 1 >0 λ 2 > 0 λ 3 > 0 28

Case 1: λ

1

=0 λ

2

= 0 λ

3

= 0

Kuhn Tucker conditions are: 𝑦 = 0 𝑥 = 0 𝜆 1 ≥ 0, 𝜆 2 ≥ 0, 𝑥 + 𝑦 ≤ 6, −𝑥 ≤ 0, 𝜆 3 ≥ 0, −𝑦 ≤ 0, 𝜆 1 𝑥 + 𝑦 − 6 = 0 𝜆 2 −𝑥 = 0 𝜆 3 −𝑦 = 0 All conditions are satisfied, but the first derivatives of the objective function, evaluated at x=y=0 are equal to zero. Then this could be a solution.

29

Consider now λ

1

=0

Kuhn Tucker conditions are: 𝜆 1 ≥ 0, 𝜆 2 𝜆 3 ≥ 0, ≥ 0, 𝑦 + 𝜆 2 𝑥 + 𝜆 3 𝑥 + 𝑦 ≤ 6, = 0 = 0 𝜆 1 −𝑥 ≤ 0, −𝑦 ≤ 0, 𝑥 + 𝑦 − 6 = 0 𝜆 2 𝜆 3 −𝑥 = 0 −𝑦 = 0 Then 𝜆 2 = −𝑦 and 𝑥 = −𝜆 3 . If 𝜆 2 ( 𝜆 3 ) is strictly positive, then y (x) is strictly negative and does not satisfy the last two conditions.

This allows us to eliminate all combinations where at least one among 𝜆 2 and 𝜆 3 λ 1 =0 and is strictly positive, then combinations 3, 5, 7 Then we have to check only the combinations 2, 4, 6, 8 30

Case 2) λ

1

>0 λ

2

= 0 λ

3

= 0

Kuhn Tucker conditions are: 𝑦 − 𝜆 1 = 0 𝑥 − 𝜆 1 = 0 𝜆 1 ≥ 0, 𝑥 + 𝑦 = 6, −𝑥 ≤ 0, −𝑦 ≤ 0, From the first 3 conditions we have that x = y = 3 and 𝜆 1 =3 These values satisfy the last conditions and the derivatives of objective function evaluated in this point are different from zero.

31

Case 4) λ

1

>0 λ

2

> 0 λ

3

= 0

Kuhn Tucker conditions are: 𝜆 1 ≥ 0, 𝑦 − 𝜆 1 + 𝜆 2 = 0 𝑥 − 𝜆 1 = 0 𝑥 + 𝑦 ≤ 6, −𝑥 = 0, 𝜆 2 −𝑦 ≤ 0 𝜆 1 𝑥 + 𝑦 − 6 = 0 −𝑥 = 0 From condition in the 4 th line we have 𝑥 = 0 , replacing in the second line we get 𝜆 1 the initial assumption of 𝜆 1 > 0 = 0 , a contradiction with 32

Case 6) λ

1

>0 λ

2

= 0 λ

3

> 0

The first two conditions are 𝑦 − 𝜆 1 = 0 𝑥 − 𝜆 1 + 𝜆 3 = 0 𝜆 3 > 0 implies 𝑦 = 0 .

Replacing it in the first line we find that 𝜆 1 = 0 , a contradiction with the initial assumption of 𝜆 1 > 0 33

Case 8) λ

1

>0 λ

2

> 0 λ

3

> 0

Kuhn Tucker conditions are: 𝑦 − 𝜆 1 + 𝜆 2 𝑥 − 𝜆 1 + 𝜆 3 𝑥 + 𝑦 = 6 = 0 = 0 𝑥 = 0 𝑦 = 0 From the last three conditions one contradiction arises Two possible solutions 1) x = 0 and y = 0 2) x = 3 and y = 3 The second one produces the higher value of the objective function, then it is the solution of the problem 34

2) Using the modified lagrangean

𝑀 𝑥, 𝑦 = 𝑥𝑦 − 𝜆 1 (𝑥 + 𝑦 − 6) Kuhn-Tucker conditions for the modified Lagrangean: 𝜆 1 𝑥 ≥ 0, 𝑦 ≥ 0 ≥ 0, 𝑦 − 𝜆 1 ≤ 0 𝑥 − 𝜆 1 ≤ 0 𝑥 + 𝑦 ≤ 6, 𝑥 𝑦 − 𝜆 1 = 0 𝑦(𝑥 − 𝜆 1 ) = 0 𝜆 1 𝑥 + 𝑦 − 6 = 0 35

Kuhn-Tucker conditions for the modified Lagrangean: 𝜆 1 𝑥 ≥ 0, 𝑦 ≥ 0 ≥ 0, 𝑦 − 𝜆 1 𝑥 − 𝜆 1 ≤ 0 𝑥 + 𝑦 ≤ 6, ≤ 0 𝑥 𝑦 − 𝜆 𝜆 1 1 = 0 𝑦(𝑥 − 𝜆 1 ) = 0 𝑥 + 𝑦 − 6 = 0 Consider a case where x=0 and y=0, then: 𝜆 1 ≥ 0, −𝜆 1 −𝜆 1 𝑥 + 𝑦 ≤ 6, ≤ 0 ≤ 0 𝜆 1 𝑥 + 𝑦 − 6 = 0 These conditions are satisfied only for 𝜆 1 = 0 Then x=0 y=0 is a candidate to the solution (the derivatives of the objective function are equal to zero in this point) 36

Consider a case where 𝑥 > 0 and 𝑦 = 0 , then: 𝜆 1 𝑥 > 0, ≥ 0, 𝜆 1 ≤ 0 𝑥 − 𝜆 𝑥 ≤ 6, 1 ≤ 0 𝜆 1 𝑥𝜆 1 = 0 𝑥 − 6 = 0 From the first condition we get 𝜆 1 = 0 Replacing 𝜆 1 = 0 in the second condition we get 𝑥 ≤ 0 A contradiction with the initial assumption 𝑥 > 0 . 37

Consider a case where 𝑥 = 0 and 𝑦 > 0 , then: Replacing these values in the second condition we get 𝜆 1 = 0 Replacing 𝜆 1 = 0 in the first condition we get 𝑦 ≤ 0 A contradiction with the initial assumption 𝑦 > 0 .

Consider the case 𝑥 > 0 and 𝑦 > 0 𝜆 1 ≥ 0, 𝑦 − 𝜆 1 = 0 𝑥 − 𝜆 1 = 0 𝑥 + 𝑦 ≤ 6, 𝜆 1 𝑥 + 𝑦 − 6 = 0 Then 𝑦 = 𝑥 = 𝜆 1 > 0 .

The last condition implies 𝑥 + 𝑦 = 6 and then 𝑥 = 𝑦 = 3 As in the procedure using the Lagrangean 38

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39